Concept

Combinatorial curvature — where it appears

The accounting that decides whether a pattern of vertices and faces fits a sphere, a plane or a hyperbolic surface, using only how many things meet at each place. In the plane it forces one over the mean face size plus one over the mean degree to be exactly a half.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

11 nets, and one accounting. Every plane net folds onto a torus when its own translations are divided out, and a torus has Euler characteristic zero — so the quotient's vertices, edges and faces satisfy n − e + f = 0 and the number of faces is not something to be counted off a drawing but e − n. Dividing through gives one over the mean face size plus one over the mean degree equal to a half, which is the same relation that forbids a plane tiling by pentagons, reached here with no geometry in it at all. It holds for every net in the table.

Every net folds onto a torus

Divide a plane net by its own translations and the quotient is a finite graph drawn on a doughnut. A doughnut has Euler characteristic zero, so the number of faces is not something to count — it is forced, and with it a relation between how many edges meet at a vertex and how many bound a face.

classification · Flat space
3 whole-number solutions: (6, 3), (4, 4), (3, 6). Every pair of whole numbers from three to 12, with the mean face size across and the mean degree down. A square in the first colour is a pair satisfying one over p plus one over q equals a half exactly — the flat case, where a periodic net is possible — and there are 3 of them: 6 and 3, 4 and 4, 3 and 6. The lighter squares above and to the left have a sum greater than a half, which is a closed polyhedron rather than a plane tiling; the ones below and to the right have a sum less than a half and belong to a surface of negative curvature. The plane is the boundary between them and it is thin.

Three answers in whole numbers

One over the face size plus one over the degree equals a half. Ask for whole numbers and there are exactly three answers, which are the three nets everybody has drawn since childhood — and the pairs on either side of them are a closed polyhedron and a plane the plane has no room for.

classification · Tilings
The same accounting, at every coordination number. One row per number of edges at a vertex. The bill a sphere charges is 2dχ; the face worth nothing is 2d/(d − 2), which is a whole number at three, four and six and is 10/3 at five; the faces that can pay are those with fewer sides than that; and the last column is every way of paying the whole bill with faces of a single size. At three edges a vertex there are three such ways and twelve pentagons is one of them. At six there are none, which is the statement that six-fold coordination belongs to the plane and to no closed surface at all.

The twelve belongs to the vertex

Twelve pentagons is read as a fact about closing a surface. It is not: it is a fact about three edges meeting at a point. Let four edges meet instead and the sphere charges eight triangles; let five meet and it charges twenty; let six meet and it cannot be paid at all.

restriction · Curvature
Every closed surface, and the two that charge nothing. The same accounting indexed by Euler characteristic rather than by genus. An orientable surface has χ = 2 − 2g, so it only ever occupies an even row; a non-orientable one has χ = 2 − k and occupies every row from one downwards. The odd rows therefore belong to surfaces that cannot be oriented and to nothing else — and the first of them, the projective plane, charges six. Six pentagons is a bill no orientable surface presents.

The surfaces a count by genus skips

A count indexed by genus steps in twelves and lands only on even numbers. A closed surface can have any characteristic at or below two, and the odd ones belong to the surfaces that cannot be oriented — where the projective plane charges six pentagons, a bill no orientable surface ever presents.

restriction · Curvature
Where the sphere and the projective plane have no net. The number of different closed nets with three bonds at every atom and faces that are pentagons and hexagons only. On the sphere, with twelve pentagons and k hexagons for k up to 12, every count has at least one net except k = 1. On the projective plane, with six pentagons and h hexagons, each count sits under the sphere count it lifts to, since every hexagon of a projective net becomes two on the sphere. The projective counts for h = 0 to 6 are 1, 0, 0, 1, 1, 3, 3, so the projective plane has no net at h = 1 or 2: two gaps where the sphere has one. Every sphere count was found by enumeration and agrees with the published one.

A gap the sphere does not have

A net of pentagons and hexagons on the projective plane must have six pentagons, and the count permits any number of hexagons. Not every number happens. Lifting each net to the sphere turns the question into one about which cages have a centre — and the answer leaves two gaps where the sphere has one.

restriction · Curvature
A centre at every other ring, and never between. Two families of closed cage, each a tube of hexagons closed at both ends by a cap of six pentagons, taken from no rings of hexagons to 8. The top row has five faces to a ring and a pentagon at each pole; the bottom row has six and a hexagon. Each box holds the cage's number of atoms with its number of hexagons beneath, and a box is drawn solid with a dot under it when the cage has a symmetry that reverses orientation and fixes nothing — a centre, which is what lets the cage halve onto the projective plane. The five-family has one at even numbers of rings and the six-family at odd ones, so their hexagon counts are 0, 10, 20, 30 … and 8, 20, 32, 44 … — two arithmetic progressions rather than two rows.

A centre at every other ring

A census cannot settle an infinite row, and the construction proposed to settle it was a tube capped at both ends, lengthened a ring at a time. Carried out, it alternates: a centre appears at every other ring and never between, the two families it permits reach two arithmetic progressions rather than a row, and the first of them opens with exactly the cage the census found could not halve.

restriction · Curvature
Every vector realised, and not at the same hexagon count. Each row is a set of faces other than hexagons whose charge — the sum of 6 − k over them — comes to twelve, which is what a closed trivalent net on the sphere must pay. Each column is a number of hexagons added to that set, and the entry is how many different solids exist with exactly those faces, found by winding up every arrangement of them into a spiral. A dash means the search found none; a question mark means the planar reader declined the row and it is not evidence either way. Every row has an entry somewhere, which is Eberhard's theorem, and the first one is at 0, 2, 3, 4 hexagons depending on the row — so the charge decides everything except the number of hexagons, and the number of hexagons is not a function of the charge.

Everything except the hexagons

Three counts of what a closed net must carry end on the same admission: an arithmetic saying what a net must charge does not say that a net exists. Eberhard's theorem says how close the charge comes to being enough, and the answer has a shape nobody would guess — it fixes every face count except the hexagons, and the hexagons are exactly the entry it cannot see.

restriction · Curvature
The fewest contacts twelve pentagons can have, by size. For every cage of pentagons and hexagons up to forty-four atoms, the number of pairs of pentagons sharing a bond. The lower line is the fewest any cage of that size achieves — 30, 24, 21, 18, 17, 15, 14, 12, 11, 10, 9, 8 — the upper line the most, and the dashed line the bound that counting edges gives: the twelve pentagons carry sixty edges between them, a contact uses two and an edge to a hexagon uses one, so the contacts cannot fall below 30 − 3h with h hexagons. The bound is attained while the hexagons are few and goes loose at five, after which each extra hexagon removes about one contact rather than three. The number of cages at each size is printed beneath, and it is the least rather than the average that the bound is about.

How close the twelve must be

The charge fixes twelve pentagons and says nothing about where they go, because it is a sum over faces and cannot see which face touches which. What it cannot see is a graph on twelve points, and the fewest edges that graph can have falls from thirty to eight over the cages a census reaches — then keeps falling at a rate that puts its first zero exactly where the truncated icosahedron is.

restriction · Curvature

Named alongside it

The objects these essays reach for when they reach for this one.

EnumerationThe Euler characteristicCrystal netPolyhedronFree actionFullereneInversion centreOrientabilityTrivalent netAutomorphismCensusClosed surface

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